Robert Hansen posted Mar 8, 2013 12:39 AM (GSC's remarks interspersed):>> lol, I really didn't think my answer would have> stumped you. >lol, your answer stumped me - and I had a short time ago posted a question expressing my bewilderment.>>Here is a simpler version. Split the> group into two groups of 6, keep the heavier group,> split that into two groups of 3 keep the heavier> group, weight (any) 2 coins from the remain group of> 3, if they weigh the same then the other coin is the> hevier one, if not then the heavier one is the> heavier one.> > It is a binary search.> > Bob Hansen> Your 'simpler version' is just plain and simply wrong. Read the puzzle more carefully.